## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 29

#### Answer

$H'(r)=\dfrac{2(r^{2}-1)^{2}(r^{2}+3r+5)}{(2r+1)^{6}}$

#### Work Step by Step

$H(r)=\dfrac{(r^{2}-1)^{3}}{(2r+1)^{5}}$ Differentiate using the quotient rule: $H'(r)=\dfrac{[(2r+1)^{5}][(r^{2}-1)^{3}]'-[(r^{2}-1)^{3}][(2r+1)^{5}]'}{[(2r+1)^{5}]^{2}}=...$ Use the chain rule to find $[(r^{2}-1)^{3}]'$ and $[(2r+1)^{5}]'$: $...=\dfrac{[(2r+1)^{5}][3(r^{2}-1)^{2}(r^{2}-1)']-[(r^{2}-1)^{3}][5(2r+1)^{4}(2r+1)']}{[(2r+1)^{5}]^{2}}=...$ $...=\dfrac{[(2r+1)^{5}][3(r^{2}-1)^{2}(2r)]-[(r^{2}-1)^{3}][5(2r+1)^{4}(2)]}{[(2r+1)^{5}]^{2}}=...$ Simplify: $...=\dfrac{6r(2r+1)^{5}(r^{2}-1)^{2}-10(2r+1)^{4}(r^{2}-1)^{3}}{(2r+1)^{10}}=...$ Take out common factors $2$, $(2r+1)^{4}$ and $(r^{2}-1)^{2}$, then continue simplifying: $...=\dfrac{2(2r+1)^{4}(r^{2}-1)^{2}[3r(2r+1)-5(r^{2}-1)]}{(2r+1)^{10}}=...$ $...=\dfrac{2(2r+1)^{4}(r^{2}-1)^{2}(6r^{2}+3r-5r^{2}+5)}{(2r+1)^{10}}=...$ $...=\dfrac{2(2r+1)^{4}(r^{2}-1)^{2}(r^{2}+3r+5)}{(2r+1)^{10}}=...$ $...=\dfrac{2(r^{2}-1)^{2}(r^{2}+3r+5)}{(2r+1)^{6}}$

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